Finding strings which contain the most 'a' letters [closed] - scala

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I've got a question if I can somehow shorten my code - I need to find what strings in my List contain the most 'a' letters. I made it as below, but I'm almost sure that I could do it shorter, but I don't know how.
So first of all I map my list to tuple, where I have (el, num) where the num is how many the 'a' contains (using foldLeft).
Then I use maxBy so I take an element that have the most 'a' letters and then I filter all elements where the num == res2._2 (because there could be more than one element having the same number of 'a' letters, so I can't stop at maxBy
My code:
def mostA(l: List[String]): List[String] = {
val res = l.map( string => (string, string.foldLeft(0)((acc,b) => if (b == 'a') acc + 1 else acc)))
val res2 = res.maxBy(x => x._2)
val res3 = res.filter(x => x._2 == res2._2).map{case (str, num) => str}
res3
}

Answer 1

With a bit of an effort, you can do it in one go:
l.foldLeft(List[String] -> 0) { case ((strs, max), s) =>
val n = s.count(_ == letter)
if (n == max) (s::strs, max) else if (n > max) (s::Nil, n) else (strs, max)
}._1.reverse

Answer 2

Your code is already pretty small and concise. However, there is a couple of improvements we can do:
def mostLetter(letter: Char)(data: List[String]): List[String] = {
val counts = data.map(s => s -> s.count(_ == letter))
val max = counts.maxBy(_._2)._2
counts.collect {
case (str, `max`) => str
}
}
Namely, we replaced the foldLeft in the String with count and we fused together the filter + map in a single collect
Also, remember the last expression is the return thus it is unnecessary to assign it to a variable to later just return it.